Find the largest Complete Subtree in a given Binary Tree
Last Updated :
11 Jul, 2025
Given a Binary Tree, the task is to find the size and also the inorder traversal of the largest Complete sub-tree in the given Binary Tree.
Complete Binary Tree - A Binary tree is a Complete Binary Tree if all levels are filled except possibly the last level and the last level has all keys as left as possible.
Note: All Perfect Binary Trees are Complete Binary trees but the reverse is not true. If a tree is not complete then it is also not a Perfect Binary Tree.
Examples:
Input:
Output:
10
8 4 9 2 10 5 1 6 3 7
Explanation: The given tree as a whole is itself a Complete Binary Tree.
Input:
Output:
4
10 45 60 70
Explanation: The below subtree is the largest subtree that satisfies the conditions of a Complete Binary Tree
Approach:
The idea is to simply traverse the tree in a bottom-up manner. During the recursion, as the process moves from the child nodes back to the parent, the sub-tree information is passed up to the parent node. This information allows the parent node to perform the Complete Tree test in constant time. Both the left and right sub-trees pass details about whether they are Perfect or Complete, along with the maximum size of the largest complete binary sub-tree found so far.
To determine if the parent sub-tree is complete, the following three cases are evaluated:
- If the left sub-tree is Perfect and the right sub-tree is Complete, and their heights are equal, then the current sub-tree (rooted at the parent) is a Complete Binary Sub-tree. Its size is equal to the sum of the sizes of the left and right sub-trees plus one (for the root).
- If the left sub-tree is Complete and the right sub-tree is Perfect, and the height of the left sub-tree is greater than the height of the right sub-tree by one, then the current sub-tree (rooted at the parent) is a Complete Binary Sub-tree. Its size is again the sum of the sizes of the left and right sub-trees plus one (for the root). However, this sub-tree cannot be Perfect, because in this case, its left child is not perfect.
- If neither of the above conditions is satisfied, the current sub-tree cannot be a Complete Binary Tree. In this case, the maximum size of a complete binary sub-tree found so far in either the left or right sub-tree is returned. Additionally, if the current sub-tree is not complete, it cannot be perfect either.
C++
// C++ program to find the largest Complete
// subtree of the given Binary Tree
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *left;
Node *right;
Node(int x) {
data = x;
left = nullptr;
right = nullptr;
}
};
// Class to store details about the subtree
class SubtreeInfo {
public:
// To store if the subtree is perfect
bool isPerfect;
// To store if the subtree is complete
bool isComplete;
// Size of the subtree
int size;
// Root of the largest complete subtree
Node *rootTree;
};
// Helper function to calculate height
// from the size of the subtree
int getHeight(int size) {
// Height is calculated using the
// formula for a perfect binary tree
return ceil(log2(size + 1));
}
// Function to find the largest complete binary subtree
SubtreeInfo findCompleteBinaryTree(Node *root) {
// Initialize the current subtree info
SubtreeInfo currTree;
// Base case: If the tree is empty
if (root == nullptr) {
currTree.isPerfect = true;
currTree.isComplete = true;
currTree.size = 0;
currTree.rootTree = nullptr;
return currTree;
}
// Recursive calls for left and right children
SubtreeInfo leftTree = findCompleteBinaryTree(root->left);
SubtreeInfo rightTree = findCompleteBinaryTree(root->right);
// CASE - 1
// If the left subtree is perfect, the right
// is complete, and their heights are equal,
// this subtree is complete
if (leftTree.isPerfect && rightTree.isComplete
&& getHeight(leftTree.size) == getHeight(rightTree.size)) {
currTree.isComplete = true;
currTree.isPerfect = rightTree.isPerfect;
currTree.size = leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 2
// If the left subtree is complete, the right
// is perfect, and the height of the left is
// greater by one, this subtree is complete
if (leftTree.isComplete && rightTree.isPerfect &&
getHeight(leftTree.size) == getHeight(rightTree.size) + 1) {
currTree.isComplete = true;
currTree.isPerfect = false;
currTree.size = leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 3
// Otherwise, this subtree is neither perfect
// nor complete. Return the largest subtree
currTree.isPerfect = false;
currTree.isComplete = false;
currTree.size = max(leftTree.size, rightTree.size);
currTree.rootTree =
(leftTree.size > rightTree.size ? leftTree.rootTree : rightTree.rootTree);
return currTree;
}
void inorderPrint(Node *root) {
if (root != nullptr) {
inorderPrint(root->left);
cout << root->data << " ";
inorderPrint(root->right);
}
}
int main() {
// Hardcoded given Binary Tree
// 50
// / \
// 30 60
// / \ / \
// 5 20 45 70
// /
// 10
Node *root = new Node(50);
root->left = new Node(30);
root->right = new Node(60);
root->left->left = new Node(5);
root->left->right = new Node(20);
root->right->left = new Node(45);
root->right->right = new Node(70);
root->right->left->left = new Node(10);
SubtreeInfo ans = findCompleteBinaryTree(root);
cout << ans.size << endl;
inorderPrint(ans.rootTree);
return 0;
}
// Java program to find the largest Complete
// subtree of the given Binary Tree
import java.util.*;
class Node {
int data;
Node left;
Node right;
Node(int x) {
data = x;
left = null;
right = null;
}
}
class SubtreeInfo {
// To store if the subtree is perfect
boolean isPerfect;
// To store if the subtree is complete
boolean isComplete;
// Size of the subtree
int size;
// Root of the largest complete subtree
Node rootTree;
}
class GfG {
// Helper function to calculate height
// from the size of the subtree
static int getHeight(int size) {
// Height is calculated using the
// formula for a perfect binary tree
return (int)Math.ceil(Math.log(size + 1)
/ Math.log(2));
}
// Function to find the largest complete binary subtree
static SubtreeInfo findCompleteBinaryTree(Node root) {
// Initialize the current subtree info
SubtreeInfo currTree = new SubtreeInfo();
// Base case: If the tree is empty
if (root == null) {
currTree.isPerfect = true;
currTree.isComplete = true;
currTree.size = 0;
currTree.rootTree = null;
return currTree;
}
// Recursive calls for left and right children
SubtreeInfo leftTree
= findCompleteBinaryTree(root.left);
SubtreeInfo rightTree
= findCompleteBinaryTree(root.right);
// CASE - 1
// If the left subtree is perfect, the right
// is complete, and their heights are equal,
// this subtree is complete
if (leftTree.isPerfect && rightTree.isComplete
&& getHeight(leftTree.size)
== getHeight(rightTree.size)) {
currTree.isComplete = true;
currTree.isPerfect = rightTree.isPerfect;
currTree.size
= leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 2
// If the left subtree is complete, the right
// is perfect, and the height of the left is
// greater by one, this subtree is complete
if (leftTree.isComplete && rightTree.isPerfect
&& getHeight(leftTree.size)
== getHeight(rightTree.size) + 1) {
currTree.isComplete = true;
currTree.isPerfect = false;
currTree.size
= leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 3
// Otherwise, this subtree is neither perfect
// nor complete. Return the largest subtree
currTree.isPerfect = false;
currTree.isComplete = false;
currTree.size
= Math.max(leftTree.size, rightTree.size);
currTree.rootTree = (leftTree.size > rightTree.size
? leftTree.rootTree
: rightTree.rootTree);
return currTree;
}
static void inorderPrint(Node root) {
if (root != null) {
inorderPrint(root.left);
System.out.print(root.data + " ");
inorderPrint(root.right);
}
}
public static void main(String[] args) {
// Hardcoded given Binary Tree
// 50
// / \
// 30 60
// / \ / \
// 5 20 45 70
// /
// 10
Node root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(5);
root.left.right = new Node(20);
root.right.left = new Node(45);
root.right.right = new Node(70);
root.right.left.left = new Node(10);
SubtreeInfo ans = findCompleteBinaryTree(root);
System.out.println(ans.size);
inorderPrint(ans.rootTree);
}
}
# Python program to find the largest Complete
# subtree of the given Binary Tree
import math
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
class SubtreeInfo:
# Initialize the subtree info
def __init__(self):
self.isPerfect = False
self.isComplete = False
self.size = 0
self.rootTree = None
# Helper function to calculate height
# from the size of the subtree
def getHeight(size):
# Height is calculated using the
# formula for a perfect binary tree
return math.ceil(math.log2(size + 1))
# Function to find the largest complete binary subtree
def findCompleteBinaryTree(root):
# Initialize the current subtree info
currTree = SubtreeInfo()
# Base case: If the tree is empty
if root is None:
currTree.isPerfect = True
currTree.isComplete = True
currTree.size = 0
currTree.rootTree = None
return currTree
# Recursive calls for left and right children
leftTree = findCompleteBinaryTree(root.left)
rightTree = findCompleteBinaryTree(root.right)
# CASE - 1
# If the left subtree is perfect, the right
# is complete, and their heights are equal,
# this subtree is complete
if (leftTree.isPerfect and rightTree.isComplete and
getHeight(leftTree.size) == getHeight(rightTree.size)):
currTree.isComplete = True
currTree.isPerfect = rightTree.isPerfect
currTree.size = leftTree.size + rightTree.size + 1
currTree.rootTree = root
return currTree
# CASE - 2
# If the left subtree is complete, the right
# is perfect, and the height of the left is
# greater by one, this subtree is complete
if (leftTree.isComplete and rightTree.isPerfect and
getHeight(leftTree.size) == getHeight(rightTree.size) + 1):
currTree.isComplete = True
currTree.isPerfect = False
currTree.size = leftTree.size + rightTree.size + 1
currTree.rootTree = root
return currTree
# CASE - 3
# Otherwise, this subtree is neither perfect
# nor complete. Return the largest subtree
currTree.isPerfect = False
currTree.isComplete = False
currTree.size = max(leftTree.size, rightTree.size)
currTree.rootTree = (leftTree.rootTree
if leftTree.size > rightTree.size
else rightTree.rootTree)
return currTree
def inorderPrint(root):
if root is not None:
inorderPrint(root.left)
print(root.data, end=" ")
inorderPrint(root.right)
if __name__ == "__main__":
# Hardcoded given Binary Tree
# 50
# / \
# 30 60
# / \ / \
# 5 20 45 70
# /
# 10
root = Node(50)
root.left = Node(30)
root.right = Node(60)
root.left.left = Node(5)
root.left.right = Node(20)
root.right.left = Node(45)
root.right.right = Node(70)
root.right.left.left = Node(10)
ans = findCompleteBinaryTree(root)
print(ans.size)
inorderPrint(ans.rootTree)
// C# program to find the largest Complete
// subtree of the given Binary Tree
using System;
class Node {
public int data;
public Node left;
public Node right;
public Node(int x) {
data = x;
left = null;
right = null;
}
}
// Class to store details about the subtree
class SubtreeInfo {
// To store if the subtree is perfect
public bool isPerfect;
// To store if the subtree is complete
public bool isComplete;
// Size of the subtree
public int size;
// Root of the largest complete subtree
public Node rootTree;
}
class GfG {
static int GetHeight(int size) {
// Height is calculated using the formula for a
// perfect binary tree
return (int)Math.Ceiling(Math.Log(size + 1)
/ Math.Log(2));
}
// Function to find the largest complete binary subtree
static SubtreeInfo FindCompleteBinaryTree(Node root) {
// Initialize the current subtree info
SubtreeInfo currTree = new SubtreeInfo();
// Base case: If the tree is empty
if (root == null) {
currTree.isPerfect = true;
currTree.isComplete = true;
currTree.size = 0;
currTree.rootTree = null;
return currTree;
}
// Recursive calls for left and right children
SubtreeInfo leftTree
= FindCompleteBinaryTree(root.left);
SubtreeInfo rightTree
= FindCompleteBinaryTree(root.right);
// CASE - 1
// If the left subtree is perfect, the right
// is complete, and their heights are equal,
// this subtree is complete
if (leftTree.isPerfect && rightTree.isComplete
&& GetHeight(leftTree.size)
== GetHeight(rightTree.size)) {
currTree.isComplete = true;
currTree.isPerfect = rightTree.isPerfect;
currTree.size
= leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 2
// If the left subtree is complete, the right
// is perfect, and the height of the left is
// greater by one, this subtree is complete
if (leftTree.isComplete && rightTree.isPerfect
&& GetHeight(leftTree.size)
== GetHeight(rightTree.size) + 1) {
currTree.isComplete = true;
currTree.isPerfect = false;
currTree.size
= leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 3
// Otherwise, this subtree is neither perfect
// nor complete. Return the largest subtree
currTree.isPerfect = false;
currTree.isComplete = false;
currTree.size
= Math.Max(leftTree.size, rightTree.size);
currTree.rootTree = (leftTree.size > rightTree.size
? leftTree.rootTree
: rightTree.rootTree);
return currTree;
}
static void InorderPrint(Node root) {
if (root != null) {
InorderPrint(root.left);
Console.Write(root.data + " ");
InorderPrint(root.right);
}
}
static void Main(string[] args) {
// Hardcoded given Binary Tree
// 50
// / \
// 30 60
// / \ / \
// 5 20 45 70
// /
// 10
Node root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(5);
root.left.right = new Node(20);
root.right.left = new Node(45);
root.right.right = new Node(70);
root.right.left.left = new Node(10);
SubtreeInfo ans = FindCompleteBinaryTree(root);
Console.WriteLine(ans.size);
InorderPrint(ans.rootTree);
}
}
// JavaScript program to find the largest Complete
// subtree of the given Binary Tree
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
class SubtreeInfo {
constructor() {
// To store if the subtree is perfect
this.isPerfect = false;
// To store if the subtree is complete
this.isComplete = false;
// Size of the subtree
this.size = 0;
// Root of the largest complete subtree
this.rootTree = null;
}
}
// Helper function to calculate height
// from the size of the subtree
function getHeight(size) {
// Height is calculated using the
// formula for a perfect binary tree
return Math.ceil(Math.log2(size + 1));
}
// Function to find the largest complete binary subtree
function findCompleteBinaryTree(root) {
// Initialize the current subtree info
let currTree = new SubtreeInfo();
// Base case: If the tree is empty
if (root === null) {
currTree.isPerfect = true;
currTree.isComplete = true;
currTree.size = 0;
currTree.rootTree = null;
return currTree;
}
// Recursive calls for left and right children
let leftTree = findCompleteBinaryTree(root.left);
let rightTree = findCompleteBinaryTree(root.right);
// CASE - 1
// If the left subtree is perfect, the right
// is complete, and their heights are equal,
// this subtree is complete
if (leftTree.isPerfect && rightTree.isComplete
&& getHeight(leftTree.size)
=== getHeight(rightTree.size)) {
currTree.isComplete = true;
currTree.isPerfect = rightTree.isPerfect;
currTree.size = leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 2
// If the left subtree is complete, the right
// is perfect, and the height of the left is
// greater by one, this subtree is complete
if (leftTree.isComplete && rightTree.isPerfect
&& getHeight(leftTree.size)
=== getHeight(rightTree.size) + 1) {
currTree.isComplete = true;
currTree.isPerfect = false;
currTree.size = leftTree.size + rightTree.size + 1;
currTree.rootTree = root;
return currTree;
}
// CASE - 3
// Otherwise, this subtree is neither perfect
// nor complete. Return the largest subtree
currTree.isPerfect = false;
currTree.isComplete = false;
currTree.size = Math.max(leftTree.size, rightTree.size);
currTree.rootTree = leftTree.size > rightTree.size
? leftTree.rootTree
: rightTree.rootTree;
return currTree;
}
function inorderPrint(root) {
let result = [];
function inorder(node) {
if (node !== null) {
inorder(node.left);
result.push(node.data);
inorder(node.right);
}
}
inorder(root);
console.log(result.join(" "));
}
// Hardcoded given Binary Tree
// 50
// / \
// 30 60
// / \ / \
// 5 20 45 70
// /
// 10
let root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(5);
root.left.right = new Node(20);
root.right.left = new Node(45);
root.right.right = new Node(70);
root.right.left.left = new Node(10);
let ans = findCompleteBinaryTree(root);
console.log(ans.size);
inorderPrint(ans.rootTree);
Time Complexity: O(n), as each node is visited once in the recursion, where n is the total number of nodes in the tree.
Auxiliary Space: O(h), due to the size of the stack used for recursion
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